mCa(OH)2 = 148*10/100=14.8g
nCa(OH)2 = 14.8/74=0.2 mol
mHCl = 36.5*20/100=7.3g
nHCl= 7.3/36.5=0.2 mol
Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
Bđ: 0.2_______0.2
Pư: 0.1_______0.2_____0.1
Kt: 0.1_________0_______0.1
mCaCl2= 0.1*111=11.1g
mdd sau phản ứng = 148 + 36.5 = 184.5g
mCa(OH)2 dư = 0.1*74= 7.4g
C%Ca(OH)2 = 7.4/184.5*100% = 4.01%
C%CaCl2 = 11.1/184.5*100%= 6.01%